best fractal software

IFS Algorithm Code function Ifs_draw (m,p) N=300000; fork=1: Length (p); eval (['a', Num2str (k),'=reshape (M (', Num2str (k),',:), 2,3);']); ENDXY=zeros (2, N); PP=Meshgrid (P); PP=Tril (PP); pp=sum (PP,2); fork=1: N-1; A=rand-pp; d=find (a0); XY (:, K+1) =eval (['a', Num2str (D (1)),'(:, 1:2)']) *xy (:, k) +eval (['a', Num2str (D (1)),'(:, 3)']); End P=complex (XY (1,:), XY (2,:)); Plot (P,'y.','Markerfacecolor','y','markersize',2); Axis equal; Axis Image;axis off;Set(GCF,'Color','k');

Poj 1941 The Sierpinski Fractal Recursion // Poj 1941 // sep9 # include Using namespace std; const int maxW = 2048; const int maxH = 1024; int pow2 [32]; char g [maxH + 10] [maxW + 10]; void print (int x, int y, int n) {if (n = 1) {g [x] [y + 1] = '/'; g [x] [y + 2] = '\'; g [x + 1] [y] = '/'; g [x + 1] [y + 1] = '_'; g [x + 1] [y + 2] = '_'; g [x + 1] [y + 3] = '\'; return;} print (x, y + pow2 [n-1], n-1 ); print (x + pow2 [n-1], y, n-1); prin

The code is as follows:ImportTurtledefMain (): T=Turtle. Turtle () T.hideturtle () T.speed (10) level= 12fract (T,-80,60,80,60, Level)deffract (t,x1,y1,x2,y2,level): Newx=0 Newy=0ifLevel = =0:drawline (t,x1,y1,x2,y2)Else: Newx= (X1+X2)/2 + (Y2-Y1)/2Newy= (y1+y2)/2-(X2-X1)/2fract (T,x1,y1,newx,newy,level-1) fract (T,newx,newy,x2,y2,level-1)defDrawLine (t,x1,y1,x2,y2): T.up () T.goto (x1,y1) T.down () T.goto (x2,y2) main ()EffectPython uses recursion to implement a

structure of one processing once again returning to the previous level to calculate again. This is a system with self-feedback. Frequent use of this feedback method makes it possible to learn and evolve the network, and this extremely simple law breeds complex phenomena. While this complex phenomenon is unpredictable, you can judge a bird's movement by taking a bird, but the movement of a flock of birds is difficult to judge, either to the left or to the right, or up or down, and one of the rea

(linker, "/stack:1024000000,1024000000")2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Set>Ten#include One#include A#include -#include -#include the using namespacestd; - intdirx[]={0,0,-1,1}; - intdiry[]={-1,1,0,0}; - #definePI ACOs (-1.0) + #defineMax (a) (a) > (b)? (a): (b) - #defineMin (a) (a) + #definell Long Long A #defineEPS 1e-10 at #defineMOD 1000000007 - #defineN 1000000 - #defineINF 1e12 - DoubleN; - intnum[100000]; - in intMain () - { to i

Dyed in three colors, colorless-0, black-1, white-2.Meet DFS Framework, very good ^.^1#include 2#include string.h>3 intCOLOR[MAXN];4 BOOLBipartite (intu) {5 for(intI=0; I)6 {7 intv=G[u][i];8 if(color[v]==0){9color[v]=3-Color[u];Ten if(!bipartite (v)) One return false; A } - Else{ - if(color[v]==Color[u]) { the return false; - } - } - } + return true; -}The staining method

Topic Links:http://poj.org/problem?id=2083Title Description:n = 1 o'clock, graph b[1] is Xn = 2 o'clock, graphic b[2] is x xXx xSo n, graphic b[n] is b[n-1] b[n-1]B[N-1]B[n-1] b[n-1]Problem Solving Ideas:The output is a rectangle, which is a rectangular!!!!!!, which is stored in a two-dimensional array with a recursive print graph.Code:1#include 2#include 3#include 4#include 5 using namespacestd;6 #defineMAXN 7407 CharMAP[MAXN][MAXN];8 9 voidDFS (intNintXinty);Ten //N is b[n], (x, y) is b[n] the

Click"Layer Properties"Window"OutsideOn the Panel, select Browse.In theStandard. ucl"Select"Triangle Inequality average"Algorithm,You can immediately seeFragmentA new color is displayed on the image, so you can take this opportunity to learn more about the gradient.In the gradient Editor, we can adjust each control point to generate a random gradient and load a pre-saved gradient.How do I load a pre-saved gradient?1. Press "F4" to open the gradient editor.2. Click"File"--"Replace", Open the" sel

Poj 1941 The Sierpinski Fractal recursion, pojsierpinski // Poj 1941 // sep9 # include Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.

If you observe"FractalProperties"Window"Layers"Panel, you will see two layers and Their thumbnail list.But when you watch yourFragmentYou can only see the display of the grayscale style layer.This is because the top layer is displayed in "normal" mode of 100%, which means that you can see any layer below it now.To check whether other layers are still in use, click"Layer 1"" Icon of the display layer, which is used to control whether the layer is displayed.In theLayers"Two Control layers are disp

Portal Fractal I still remember that this question was the first week of the competition ~~~ Why can't I cut it out ~~ At that time, I tried againCode~~ Still do not understand Today, I think that the idea of divide, conquer, and Recursion really solves this problem that has plagued me for a long time. The original Article is like this. X-XXX-XXXXXXXXX This is the idea of recursive extra points. Example 1: simple Example 2: Add x

Another way to view multiple layers at a time is to set them in fusion mode.This setting determines the combination of each pixel in the layer and each pixel in the other layers below.First,Layer 1"The opacity of the layer is reset to 100%, click the drop-down box in fusion mode, and select the next option --"Multiply".In this way, we canLayer 1"Layer to see the background layer (although it looks dark ).The display effect is as follows:The arrow keys allow us to switch to the fusion mode. Remem

　　Topic Link: http://hihocoder.com/problemset/problem/1122, Hungarian algorithm naked question.Just learn the two points to match, or to more brush questions. This problem can be directly set template, I am based on the topic above to do, so first added a dyeing optimization, the effect of the general bar.#include #include#include#include#include#includestring>#includestring.h>#includeusing namespacestd;#defineLL __int64#defineEPS 1e-8#defineINF 1e8#defineLson L, M, RT #defineRson m + 1, R

)/3, A.Y + (B.Y-A.Y)/3);Point d = new Point (a.x + 2 * (b.x-a.x)/3, A.Y + 2 * (B.Y-A.Y)/3);Double L = math.sqrt ((c.x-d.x) * (c.x-d.x) + (C.Y-D.Y) * (C.Y-D.Y));Point e;if (b.y-a.y >= 0 b.x-a.x > 0){v = r;}else if (b.y-a.y > 0 b.x-a.x {v = math.pi-r;}else if (b.y-a.y {v = Math.PI + R;}else if (B.y-a.y {v = 2 * math.pi-r;}E = new Point ((float) (L * Math.Cos (v + math.pi/3) + c.x), (float) (c.y + L * Math.sin (v + math.pi/3)));Line (A, C, n-1);Line (c, E, n-1);Line (E, D, n-1);Line (d, b, n-1);}

Any object in a ray tracing scenario can be rendered only by the intersection algorithm and the normal algorithm.Virtual int intersect (const ray ray, float distance) = 0; Virtual vector3 getnormal (vector3 Pos) = 0; To adapt to the rendering of the triangle band, the data structure of the triangle is defined: Vector3 vertex [3];Vector3 normal [3]; Because pasters are not yet supported, the ray tracing plug-in only processes vertex data (0x4110) and surface data (0x4120) when reading the

"Data Recovery failure description"Company financial personnel for data maintenance, misoperation, in the financial software to initialize the data, because recently did not do backup, it caused a lot of financial documents lost.Because the financial data is very important, the customer is anxious to get back.The data Recovery Analysis "Engineer detection, in SQL Server database Enterprise Manager, discover what the

Created in the previous sectionFragmentThe pattern is not satisfactory.ArtProductBut before we proceed, let's save our files first.There are multiple ways to save it. We will introduce it in the next tutorial.This time, we save the image made in UF

Software can be roughly divided into: Free Software and non-free software Types of Free Software and non-free software.The following are some terms that are frequently mentioned when discussing free software. They explain which types overlap with others or are part of o

It takes more than two years to develop programs! Make a small summary! Status Quo, At present, many software companies in China are still working in small workshops! After receiving a project, let's discuss how to do it. After the boss decides to make a decision, let's split our heads! As a result, a project started, and everyone was busy in the early stage! Everyone has their own jobs (some companies are unevenly allocated. If there are idle people,

Software Engineering-software testing, Software Engineering Software Testing After the system is complete, it is time to test the system. This must have been the first system of acceptance. We are confident that the system will be accepted by the master. Once it comes to the master, nothing will happen, in addition, so